What is the sum of the squares of the numbers from 3 to 18?

Correct option is B

Formula Used:

Sum of the squares of first n natural numbers =$$\frac{n(n+1)(2n+1)}{6}$$

Solution:

Sum of square of first 18 numbers = $$\frac{18(18+1)(36+1)}{6}$$= 2109

Sum of square of first 2 numbers=$$\frac{2(2+1)(4+1)}{6}$$ =5

Sum of square from 3 to 18 = 2109 -5 =2104